The definition of vector derivatives I am reading a textbook about the derivatives, and it says..
Let $x, y, f(x)\in \mathbb{R}$ then we can define $f'(x)$ as the following:  
$$\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}=f'(x)$$ 
However if $x$ is a vector, then equivalent to the above is the condition  
$$f(x+h) = f(x)+f'(x)h+R(h) \Rightarrow \lim_{h\rightarrow 0}\frac{R(h)}{|h|}=0$$  
I am quite confused about why is this equivalent and what is $R(h)$ here?  Please advise, thanks!
 A: Think of the existence of the derivative as saying the function is “approximately linear” at that point. For the one variable case say we want to approximate the value of $f$ at a point near $a$, say $x$, using the value of $f(a)$. We can write $f(x)= f(a) + f’(a) h + R(h)$ where we took $x=a+h$. Do you see how this says $f$ is locally like a line given that the remainder $R$ gets sufficiently small? That is, $R(h)/h \to 0$ as $h \to 0$. The same thing happens in higher dimensions, except we can’t express it as a quotient because we’re dealing with vectors. Instead we express that the remainder (which is now a vector) has a magnitude that gets sufficiently small so the function $f$ is “locally linear”. Or $\|R(h)\| / |h| \to 0$
A: In your formula, it is missing that the limit of $\frac{R(h)}{|h|}$ should be 0. 
R(h) stands for the (order one) Taylor reminder, which is defined implicitly by the first equation i.e.:
$$ R(h)= f(x+h) - [f(x)+ f^\prime(x) h] $$
Geometrically, it measures the error in the approximation to the graph of f by the tangent line at x.
